When excess of ammonia is used:
$3 C l_{2} + 8 N H_{3}$ $⟶$ $A + B$
When excess of chlorine is used:
$3 C l_{2} + 4 N H_{3}$ $⟶$ $C + D$
Oxidation state of nitrogen in compound $A$ is $a$ and Oxidation state of nitrogen in compound $D$ is $d$ . Value of $a - d$ is:

- 6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 6$
$8 N H_{3} + 3 C l_{2} \to 6 N H_{4} C l + N_{2}$

$3 C l_{2} + 4 N H_{3} \to 3 N H_{4} C l + N C l_{3}$
Compound A is $N H_{4} C l$ and compound D is $N C l_{3}$
So the oxidation state of nitrogen in $N H_{4} C l$ is $- 3$
the oxidation state of nitrogen in $N C l_{3}$ is $+ 3$
So $a - d$ is $- 3 - (+ 3)$
$- 3 - 3$ that is equals to $- 6$
Hence option A is correct.

Suggest Corrections

Similar questions

Q. When excess of ammonia is used:

3 C l_{2} + 8 N H_{3}

⟶

A + B

When excess of chlorine is used:

3 C l_{2} + 4 N H_{3}

⟶

C + D

The oxidation state of nitrogen in compound

D

is:

A mole of $N_{2} H_{4}$ loses 10 mol of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation number of hydrogen.)

Question 15
Ozone-layer is getting depleted because of

(a) excessive use of automobiles
(b) excessive formation of industrial units
(c) excessive use of man-made compounds containing both fluorine and chlorine
(d) excessive deforestation.

Question 15
Ozone-layer is getting depleted because of

(a) excessive use of automobiles
(b) excessive formation of industrial units
(c) excessive use of man-made compounds containing both fluorine and chlorine
(d) excessive deforestation.

Q. The compound of nitrogen in which nitrogen carries a fractional oxidation state is ___________.

Join BYJU'S Learning Program

When excess of ammonia is used: 3Cl2+8NH3⟶A+BWhen excess of chlorine is used:3Cl2+4NH3⟶C+DOxidation state of nitrogen in compound A is a and Oxidation state of nitrogen in compound D is d. Value of a−d is:

The correct option is A −68NH3+3Cl2→6NH4Cl+N2 3Cl2+4NH3→3NH4Cl+NCl3Compound A is NH4Cl and compound D is NCl3So the oxidation state of nitrogen in NH4Cl is −3the oxidation state of nitrogen in NCl3 is +3So a−d is −3−(+3)−3−3 that is equals to −6Hence option A is correct.

When excess of ammonia is used:
$3 C l_{2} + 8 N H_{3}$ $⟶$ $A + B$
When excess of chlorine is used:
$3 C l_{2} + 4 N H_{3}$ $⟶$ $C + D$
Oxidation state of nitrogen in compound $A$ is $a$ and Oxidation state of nitrogen in compound $D$ is $d$ . Value of $a - d$ is: